Let $X$ be a topological vector space (no assumptions about local convexity are made in the question, though I am worried they might be required). Suppose $K\subset X$ is a compact, convex, metrizable subset of $X$, and denote by $\partial_e K$ the extreme boundary of $K$. We wish to show that $\partial_e K$ is $G_\delta$ in $K$.

It seems that this would boil down to writing a correct, alternate definition of being an extreme point as an intersection of bunch of open or $G_\delta$ sets, and using separability (of $K$ or of $\mathbb{R}$) to reduce that intersection to a countable one. I had two possible approaches:

(2): Let $D$ be a countable dense set in $K$ (which is separable). For $y\neq z$ in $D$, let $$G_{y,z}=\{x\in K:x \text{ is not in the interior of the line segment between $y$ and $z$}\}.$$ It is easy to see that each $G_{y,z}$ is $G_\delta$, and $\partial_e K\subset \bigcap_{y\neq z \in D} G_{y,z}$. There is an obvious problem if $D$ is dense in the interior of $K$ while containing no boundary points of $K$, but we could throw in a countable dense subset of the boundary of $K$ as well. Under the assumption of local convexity, we need only show that every boundary point of $K$ which is in $\bigcap_{y\neq z \in D} G_{y,z}$ is in $\partial_e K$, which seems geometrically clear to me, but I do not see the details.

1 Answer
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Let $K$ be compact, convex and metrizable in the topological vector space $X$ (local convexity isn't required for the argument). Fix a metric $d$ on $K$ compatible with the topology. Let $$F_n = \left\{x \in K\,:\, \text{there are } y,z \in K\text{ such that }x = \frac{1}{2}(y+z)\text{ and }d(y,z) \geq \frac{1}{n}\right\}.$$
Then $F_n$ is closed and a point is non-extremal if and only if it is in the $F_\sigma$-set $F = \bigcup_n F_n$. Thus the set of extremal points $\partial_e K = K \smallsetminus F$ is a $G_\delta$.

The set $C_n = \{(y,z) \in K \times K\,:\,d(y,z) \geq 1/n\}$ is closed since it is the preimage of $[1/n,\infty)$ under the continuous function $d\colon K \times K \to [0,\infty)$. Therefore $C_n$ is compact. The set $F_n$ is the image of the compact set $C_n$ under the continuous function $(y,z) \mapsto \frac{1}{2}(y+z)$, hence $F_n$ is compact and thus closed.

Conversely, if $x \in F_n$ for some $n$ then $x$ is clearly not extremal.

Later:

Let me add some remarks on your approaches:

The problem with idea (1) is that $U_{\lambda}$ is not open. In fact, I showed in point 2. above that $U_{1/2} = \partial_e K$ and a small modification of that argument gives that $U_{\lambda} = \partial_e K$ for all $\lambda \in (0,1)$, so you're right that $\partial_e K = \bigcap_{\lambda} U_{\lambda}$, but proving that $U_{\lambda}$ is a $G_\delta$ is the same as the original problem, so that idea won't lead anywhere.

The second idea looks much better, however I doubt that exploiting separability of $K$ only is enough (that is: I doubt that the set of extremal points of a compact convex separable but non-metrizable set is a $G_\delta$, but I haven't checked this thoroughly). I think the argument I gave is one way to get around the difficulties.

Another point I'd like to add is that the distinction of boundary points and interior points you seem to be making does not work for infinite-dimensional compact convex sets. In fact, the Hilbert cube $C$ is homogeneous in the sense that its homeomorphism group acts transitively (see here for a good write-up of the non-trivial proof), so no point is distinguished by topological properties alone. This property is generic in the sense that every compact convex metrizable set in a locally convex vector space is either contained in a finite-dimensional subspace or homeomorphic to $C$ by a theorem of Klee. See my answer here for more on this.

Finally, we can't hope to do much better than $G_\delta$. In three dimensions take the double cone $K$ obtained by taking the convex hull of a circle $C$ of radius $1$ around $(1,0,0)$ in the $(x,y)$-plane and the two points $p_\pm=(0,0,\pm1)$ on the $z$-axis. Then $\partial_e K = \{p_\pm\} \cup C\smallsetminus \{(0,0,0)\}$ is not closed. [In two dimensions the non-extremal points are open in the boundary, hence the set of extremal points of a compact convex set is closed, the one-dimensional case is trivial.]

I do not remember where I learned this argument, most likely in Phelps's Lectures on Choquet's theorem (also mentioned in Pietro Majer's answer on MO).
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t.b.Jun 12 '12 at 3:41

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Congrats to your 500-th answer, t.b. BTW this is also given as Exercise 4.10 in Kechris' Descriptive Set Theory; but I don't remember what was my solution when I was reading that chapter.
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Martin SleziakJun 12 '12 at 6:13

@Martin: Thanks and thanks for the reference! Given this recent question Kechris is very likely the source of this problem :)
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t.b.Jun 12 '12 at 6:28

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Thanks for the answer as well as the other info! Also, you are correct that this problem is sourced from the exercise in Kechris.
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Iian SmytheJun 12 '12 at 15:50